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## Homework Statement

Consider a system in which R 134-A is moved through an evaporator. The evaporator removes energy from the the surroundings at a rate of 0.54 kW. The pressure = 120 kPa and x = 0.2 right before the fluid enters the evaporator. After exiting the evaporator, the pressure = 120 kPa and T = -20 C. What is the mass flow rate through the evaporator?

## Homework Equations

(1) [tex]\sum[/tex]mass_in = [tex]\sum[/tex]mass_out

(2) [tex]\dot{q}[/tex] - [tex]\dot{w}[/tex] = [tex]\sum[/tex](([tex]\dot{m}[/tex]_out) * (h + [(V_2)^2/2] + gz)) - [tex]\sum[/tex](([tex]\dot{m}[/tex]_in) * (h + [(V_1)^2/2] + gz))

(3) [tex]\dot{E}[/tex] = [tex]\dot{m}[/tex]e

where e = u + ke + pe

## The Attempt at a Solution

Assuming steady flow, equation 1 is applicable, so the mass flow rate going into the evaporator is the same as the mass flow rate exiting the evaporator. However, I'm not exactly sure how to go about from there.